Unformatted text preview: then
those without both John and Pat
8
7
8
6.5.4 +
5.4 +
6.5.4 = 10, 780
5
4
6
How many anagrams of MISSISSPPI?
11!/(4!4!2!1!) = 34, 650
Math 30530 (Fall 2012) Counting September 13, 2013 6 / 12 Some properties of binomial coefﬁcients
n
n = 1 (justiﬁes 0! = 1) and Math 30530 (Fall 2012) n
0 =1 Counting September 13, 2013 7 / 12 Some properties of binomial coefﬁcients
n
n = 1 (justiﬁes 0! = 1) and n
k = n
0 =1 n
n−k Math 30530 (Fall 2012) Counting September 13, 2013 7 / 12 Some properties of binomial coefﬁcients
n
n = 1 (justiﬁes 0! = 1) and n
k = n
0 =1 n
n−k (x + y )n = Math 30530 (Fall 2012) n
n
k =0 k x k y n−k (Binomial Theorem) Counting September 13, 2013 7 / 12 Some properties of binomial coefﬁcients
n
n = 1 (justiﬁes 0! = 1) and n
k = n
0 =1 n
n−k (x + y )n =
Special case: Math 30530 (Fall 2012) n
n
k =0 k x k y n−k (Binomial Theorem) n
n
k =0 k = 2n Counting September 13, 2013 7 / 12 Some properties of binomial coefﬁcients
n
n = 1 (justiﬁes 0! = 1) and n
k = n
0 =1 n
n−k (x + y )n =
Special case: n
n
k =0 k x k y n−k (Binomial Theorem) n
n
k =0 k = 2n If I repeat the same trial n times, independently, and each time I
have probability p of success, then the probability that I have
n
exactly k successes is k pk (1 − p)n−k Math 30530 (Fall 2012) Counting September 13, 2013 7 / 12 Some properties of binomial coefﬁcients
n
n = 1 (justiﬁes 0! = 1) and n
k = n
0 =1 n
n−k (x + y )n =
Special case: n
n
k =0 k x k y n−k (Binomial Theorem) n
n
k =0 k = 2n If I repeat the same trial n times, independently, and each time I
have probability p of success, then the probability that I have
n
exactly k successes is k pk (1 − p)n−k
Many identities, e.g.
Committee/Chair: n Math 30530 (Fall 2012) n−1
k −1 =k n
k Counting September 13, 2013 7 / 12 Some properties of binomial coefﬁcients
n
n = 1 (justiﬁes 0! = 1) and n
k = n
0 =1 n
n−k
n
n
k =0 k (x + y )n =
Special case: x k y n−k (Binomial Theorem) n
n
k =0 k = 2n If I repeat the same trial n times, independently, and each time I
have probability p of success, then the probability that I have
n
exactly k successes is k pk (1 − p)n−k
Many identities, e.g.
n−1
k −1 =
n −1
n −1
k −1 +
k Committee/Chair: n
Pascal: Math 30530 (Fall 2012) n
k = k n
k...
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This note was uploaded on 01/31/2014 for the course MATH 30530 taught by Professor Hind,r during the Fall '08 term at Notre Dame.
 Fall '08
 Hind,R
 Counting, Probability

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